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Re: Transatlantic response times.
On Mon, Mar 25, 2002 at 09:13:20AM -0600, Pistone, Mike wrote: > Is there any equation to estimate response times? For example, if your > circuit from A to Z has a 500ms avg response, than that equates to a circuit > distance of aprox. 5000 miles or something? As I'm sure you remember from your physics class, light travels through a vacuum at 299,792 km/sec (lets call it 300,000). When it travels through other mediums, it moves slower based on that medium's refractive index. For example, water has a refractive index of 1.33, which means light travels through water at about 0.75c, or about 225,000 km/sec. Fiber works on a principal called "total internal refraction", which means that the light is continually reflected into the core with no (or little) loss in the cladding. To accomplish this, different material with different refractive indexes is used. Since the cladding has a lower refractive index than the core, as long as the angle of incidence exceeds a critical angle, the light will be reflected back into the core instead of shooting out the sides. The values of refractive indexes used in fiber are usually something along the lines of 1.46 in the cladding and 1.48 in the core. So if you do a little math, you'll see that light propagates through fiber at around 0.67c, or 200,000 km/sec (or approx 125,000 miles/sec). Putting that in ms terms so even ping monkeys understand, you get approximately 1ms of speed-of-light induced delay per 200 km (or 125 miles) of fiber path. As an example (and to answer your original question), 500ms RTT / 2 is approximately a 50,000 km or 31,250 mile fiber path. Adjust a little for all the microseconds of switching and buffering which happens to your packet along the way, and you can get a fairly good idea how drunk the people were when they laid your fiber. -- Richard A Steenbergen <[email protected]> http://www.e-gerbil.net/ras PGP Key ID: 0x138EA177 (67 29 D7 BC E8 18 3E DA B2 46 B3 D8 14 36 FE B6)